Show that $\mathbb{E}[ | X - \mathbb{E}[X] | ] \ge c > 0$ if $\mathbb{V}(X) = 1$? I have a random variable $X$ with unit variance, i.e. $\mathbb{V}(X) = 1$. Is there a positive constant $c > 0$ such that
$$\mathbb{E}[\ | X - \mathbb{E}[X] | \ ] \ge c > 0  ?$$
How do we prove it? If it is false, can we find a counter-example?
 A: I will go ahead and formalise the answer in the comments from Glen_b.  For any value $0 < \epsilon \leqslant \tfrac{1}{2}$ we can define a discrete random variable $X$ with mass function:
$$\mathbb{P}(X=x)
= \begin{cases} 
\epsilon & & & \text{for } x = -\tfrac{1}{\sqrt{2 \epsilon}}, \\[6pt]
1-2\epsilon & & & \text{for } x = 0, \\[10pt]
\epsilon & & & \text{for } x = \tfrac{1}{\sqrt{2 \epsilon}}, \\[10pt]
0 & & & \text{otherwise}. \\[6pt]
\end{cases}$$
This random variable has zero mean and has variance:
$$\begin{align}
\mathbb{V}(X)
= \mathbb{E}(X^2)
&= (-\tfrac{1}{\sqrt{2 \epsilon}})^2 \cdot \epsilon + 0 \cdot (1-2\epsilon) + (\tfrac{1}{\sqrt{2 \epsilon}})^2 \cdot \epsilon \\[6pt]
&= \tfrac{1}{2 \epsilon} \cdot \epsilon + \tfrac{1}{2 \epsilon} \cdot \epsilon \\[8pt]
&= \tfrac{1}{2} + \tfrac{1}{2} = 1. \\[6pt]
\end{align}$$
For this random variable we have:
$$\begin{align}
\mathbb{P}(|X - \mathbb{E}(X)|)
&= \mathbb{P}(|X|) \\[8pt]
&= |-\tfrac{1}{\sqrt{2 \epsilon}}| \cdot \epsilon + |0| \cdot (1-2\epsilon) + |\tfrac{1}{\sqrt{2 \epsilon}}| \cdot \epsilon \\[6pt]
&= \tfrac{1}{\sqrt{2 \epsilon}} \cdot \epsilon + \tfrac{1}{\sqrt{2 \epsilon}} \cdot \epsilon \\[6pt]
&= \sqrt{\tfrac{\epsilon}{2}} + \sqrt{\tfrac{\epsilon}{2}} \\[6pt]
&= \sqrt{2\epsilon}. \\[6pt]
\end{align}$$
Now, since we can use any value $0 < \epsilon \leqslant \tfrac{1}{2}$ for this analysis, there is no positive lower bound for this probability --- i.e., its lower bound is zero.
A: Let $Y = X-\mathbb{E}(X)$. Then by Chebyshev's inequality, for any $k > 0$, $P(|Y-\mathbb{E}[Y] | \geq k) \leq 1/k^{2} \geq P(|Y| \geq k)$. By Chebyshev's inequality, think you can reach a contradiction.
